Line Questions in English

(C) The vector equation of a line passing through two points with position vectors $a$ and $b$ is given by $r = a + t(b - a)$,where $t$ is a scalar parameter.
Here,$a = i - 2j + k$ and $b = -2j + 3k$.
Calculating the direction vector $(b - a)$:
$b - a = (-2j + 3k) - (i - 2j + k) = -i + 0j + 2k = 2k - i$.
Substituting these into the formula:
$r = (i - 2j + k) + t(2k - i)$.

(A) Let $A$ be the point $(-1, 2, 6)$ and $P$ be the point $(2, 3, -4)$ on the line. The vector $\overrightarrow{PA}$ is given by $\overrightarrow{A} - \overrightarrow{P} = (-1-2)i + (2-3)j + (6-(-4))k = -3i - j + 10k$.
The line is parallel to the vector $\vec{v} = 6i + 3j - 4k$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
The projection of $\overrightarrow{PA}$ on the line is given by $PN = \left| \frac{\overrightarrow{PA} \cdot \vec{v}}{|\vec{v}|} \right| = \left| \frac{(-3)(6) + (-1)(3) + (10)(-4)}{\sqrt{61}} \right| = \left| \frac{-18 - 3 - 40}{\sqrt{61}} \right| = \left| \frac{-61}{\sqrt{61}} \right| = \sqrt{61}$.
In the right-angled triangle $\triangle APN$,the distance $AN$ from point $A$ to the line is given by $AN = \sqrt{|\overrightarrow{PA}|^2 - PN^2}$.
We have $|\overrightarrow{PA}|^2 = (-3)^2 + (-1)^2 + 10^2 = 9 + 1 + 100 = 110$.
Thus,$AN = \sqrt{110 - 61} = \sqrt{49} = 7$.

(C) The given lines are in the form $r = a_1 + t b_1$ and $r = a_2 + s b_2$,where:
$a_1 = 3i - 2j - 2k$,$b_1 = i$
$a_2 = i - j + 2k$,$b_2 = j$
First,calculate the cross product of the direction vectors:
$b_1 \times b_2 = i \times j = k$
$|b_1 \times b_2| = |k| = 1$
Next,calculate the vector $(a_2 - a_1)$:
$a_2 - a_1 = (i - j + 2k) - (3i - 2j - 2k) = -2i + j + 4k$
The shortest distance $d$ is given by the formula:
$d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|}$
Substitute the values:
$d = \frac{|(-2i + j + 4k) \cdot k|}{1} = \frac{|4|}{1} = 4$
Thus,the shortest distance is $4$.

(C) The vector equation of a line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by $\vec{r} = \vec{a} + t(\vec{b} - \vec{a})$,where $t$ is a scalar parameter.
Expanding this,we get $\vec{r} = \vec{a} + t\vec{b} - t\vec{a} = (1 - t)\vec{a} + t\vec{b}$.
Substituting $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ into the equation:
$\vec{r} = (1 - t)(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) + t(b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$
$\vec{r} = a_1(1 - t)\hat{i} + a_2(1 - t)\hat{j} + a_3(1 - t)\hat{k} + t(b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$.
Thus,option $C$ is correct.

(D) The given lines are in the form ${r_1} = {a_1} + \lambda {b_1}$ and ${r_2} = {a_2} + \mu {b_2}$.
Here,${a_1} = 4i - 3j - k$,${b_1} = i - 4j + 7k$,${a_2} = i - j - 10k$,and ${b_2} = 2i - 3j + 8k$.
First,we calculate the cross product ${b_1} \times {b_2}$:
${b_1} \times {b_2} = \begin{vmatrix} i & j & k \\ 1 & -4 & 7 \\ 2 & -3 & 8 \end{vmatrix} = i(-32 + 21) - j(8 - 14) + k(-3 + 8) = -11i + 6j + 5k$.
Next,we find the vector ${a_2} - {a_1} = (1-4)i + (-1 - (-3))j + (-10 - (-1))k = -3i + 2j - 9k$.
The shortest distance is given by $d = \frac{|({a_2} - {a_1}) \cdot ({b_1} \times {b_2})|}{|{b_1} \times {b_2}|}$.
Calculating the dot product: $({a_2} - {a_1}) \cdot ({b_1} \times {b_2}) = (-3)(-11) + (2)(6) + (-9)(5) = 33 + 12 - 45 = 0$.
Since the scalar triple product is $0$,the lines are coplanar and intersect,so the shortest distance is $0$.

(B) Let the ratio in which the $YOZ$ plane divides the line segment $PQ$ be $k:1$.
The coordinates of point $P$ are $(9, -1, 5)$ and point $Q$ are $(1, 3, 5)$.
The $YOZ$ plane is defined by the equation $x = 0$.
Using the section formula,the $x$-coordinate of the point $R$ dividing $PQ$ in the ratio $k:1$ is given by:
$x = \frac{k(x_2) + 1(x_1)}{k+1} = 0$
Substituting the values:
$\frac{k(1) + 1(9)}{k+1} = 0$
$k + 9 = 0$
$k = -9$
Thus,the ratio $PR : RQ$ is $-9 : 1$.

(B) Let the line pass through $A(4, 2, 2)$ and be parallel to $\vec{c} = 2i + 3j + 6k$.
Let $M$ be the foot of the perpendicular from $B(1, 2, 3)$ to the line.
Vector $\vec{AB} = (1-4)i + (2-2)j + (3-2)k = -3i + 0j + k$.
The length $AM$ is the projection of $\vec{AB}$ on the vector $\vec{c}$.
$AM = \frac{|\vec{AB} \cdot \vec{c}|}{|\vec{c}|} = \frac{|(-3)(2) + (0)(3) + (1)(6)|}{\sqrt{2^2 + 3^2 + 6^2}} = \frac{|-6 + 0 + 6|}{\sqrt{4 + 9 + 36}} = \frac{0}{7} = 0$.
In the right-angled triangle $\triangle ABM$,$BM^2 = AB^2 - AM^2$.
$AB^2 = |\vec{AB}|^2 = (-3)^2 + 0^2 + 1^2 = 9 + 0 + 1 = 10$.
$BM^2 = 10 - 0^2 = 10$.
Therefore,$BM = \sqrt{10}$.

(B) Let the $xy$-plane divide the line segment joining the points $A(2, 4, 5)$ and $B(-4, 3, -2)$ in the ratio $k:1$ at point $P$.
Since the point $P$ lies on the $xy$-plane,its $z$-coordinate must be $0$.
Using the section formula for the $z$-coordinate:
$z = \frac{k(z_2) + 1(z_1)}{k+1} = 0$
Substituting the values:
$\frac{k(-2) + 1(5)}{k+1} = 0$
$-2k + 5 = 0$
$2k = 5$
$k = \frac{5}{2}$
Thus,the ratio is $5:2$.

(B) The equation of the line passing through $(x_1, y_1, z_1) = (3, 4, 1)$ and $(x_2, y_2, z_2) = (5, 1, 6)$ is given by the section formula. Any point on this line can be represented as $\left( \frac{5\lambda + 3}{\lambda + 1}, \frac{1\lambda + 4}{\lambda + 1}, \frac{6\lambda + 1}{\lambda + 1} \right)$.
Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$.
Setting the $z$-coordinate to $0$: $\frac{6\lambda + 1}{\lambda + 1} = 0 \Rightarrow 6\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{6}$.
Substituting $\lambda = -\frac{1}{6}$ into the coordinates:
$x = \frac{5(-\frac{1}{6}) + 3}{-\frac{1}{6} + 1} = \frac{-\frac{5}{6} + 3}{\frac{5}{6}} = \frac{13/6}{5/6} = \frac{13}{5}$.
$y = \frac{1(-\frac{1}{6}) + 4}{-\frac{1}{6} + 1} = \frac{-\frac{1}{6} + 4}{\frac{5}{6}} = \frac{23/6}{5/6} = \frac{23}{5}$.
Thus,the point of intersection is $\left( \frac{13}{5}, \frac{23}{5}, 0 \right)$.

(A) The direction ratios of the line joining the points $P(-2, 1, -8)$ and $Q(a, b, c)$ are given by $(a - (-2), b - 1, c - (-8))$,which simplifies to $(a + 2, b - 1, c + 8)$.
Since this line is parallel to the line with direction ratios $6, 2, 3$,their direction ratios must be proportional.
Therefore,$\frac{a + 2}{6} = \frac{b - 1}{2} = \frac{c + 8}{3} = k$ for some constant $k$.
If we take $k = 1$,we get $a + 2 = 6 \Rightarrow a = 4$,$b - 1 = 2 \Rightarrow b = 3$,and $c + 8 = 3 \Rightarrow c = -5$.
Thus,the values are $a = 4, b = 3, c = -5$.

(A) The equation of the $yz$-plane is $x = 0$.
Let the point $P$ divide the line segment joining $A(3, 5, -7)$ and $B(-2, 1, 8)$ in the ratio $k:1$.
The coordinates of $P$ are given by $\left(\frac{-2k + 3}{k+1}, \frac{k + 5}{k+1}, \frac{8k - 7}{k+1}\right)$.
Since $P$ lies on the $yz$-plane,its $x$-coordinate must be $0$:
$\frac{-2k + 3}{k+1} = 0 \implies -2k + 3 = 0 \implies k = \frac{3}{2}$.
Substituting $k = \frac{3}{2}$ into the coordinates of $P$:
$y = \frac{\frac{3}{2} + 5}{\frac{3}{2} + 1} = \frac{\frac{13}{2}}{\frac{5}{2}} = \frac{13}{5}$.
$z = \frac{8(\frac{3}{2}) - 7}{\frac{3}{2} + 1} = \frac{12 - 7}{\frac{5}{2}} = \frac{5}{\frac{5}{2}} = 2$.
Thus,the coordinates are $\left(0, \frac{13}{5}, 2\right)$.

(A) Let $P = ({x_1}, {y_1}, {z_1})$ be the point and $A = ({x_2}, {y_2}, {z_2})$ be a point on the line. The vector $\vec{AP} = ({x_1} - {x_2})\hat{i} + ({y_1} - {y_2})\hat{j} + ({z_1} - {z_2})\hat{k}$.
Let $\vec{u} = l\hat{i} + m\hat{j} + n\hat{k}$ be the unit vector along the line.
The projection of $\vec{AP}$ on the line is $p = |\vec{AP} \cdot \vec{u}| = |l({x_1} - {x_2}) + m({y_1} - {y_2}) + n({z_1} - {z_2})|$.
By the Pythagorean theorem,the perpendicular distance $d$ is given by $d^2 = |\vec{AP}|^2 - p^2$.
$|\vec{AP}|^2 = ({x_1} - {x_2})^2 + ({y_1} - {y_2})^2 + ({z_1} - {z_2})^2$.
Thus,$d = \sqrt{({x_1} - {x_2})^2 + ({y_1} - {y_2})^2 + ({z_1} - {z_2})^2 - [l({x_1} - {x_2}) + m({y_1} - {y_2}) + n({z_1} - {z_2})]^2}$.

(D) The direction ratios of line $AB$ are $(4-1, 5-2, 7-3) = (3, 3, 4)$.
The direction ratios of line $CD$ are $(2-(-4), 9-3, 2-(-6)) = (6, 6, 8)$.
Let the angle between the lines be $\theta$. The formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Here,$a_1=3, b_1=3, c_1=4$ and $a_2=6, b_2=6, c_2=8$.
Note that $(a_2, b_2, c_2) = 2(a_1, b_1, c_1)$,which means the lines are parallel.
Therefore,$\cos \theta = \frac{|(3)(6) + (3)(6) + (4)(8)|}{\sqrt{3^2+3^2+4^2} \sqrt{6^2+6^2+8^2}} = \frac{|18+18+32|}{\sqrt{34} \sqrt{136}} = \frac{68}{\sqrt{34} \cdot 2\sqrt{34}} = \frac{68}{2 \cdot 34} = 1$.
Since $\cos \theta = 1$,the angle $\theta = 0^\circ$ or $0$ radians.
Since $0$ is not among the options $A, B, C$,the correct answer is $D$.

(A) The direction ratios of the first line are $\vec{a} = (1, 0, -1)$.
The direction ratios of the second line are $\vec{b} = (3, 4, 5)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(1)(3) + (0)(4) + (-1)(5)|}{\sqrt{1^2 + 0^2 + (-1)^2} \sqrt{3^2 + 4^2 + 5^2}}$
$\cos \theta = \frac{|3 + 0 - 5|}{\sqrt{1 + 0 + 1} \sqrt{9 + 16 + 25}}$
$\cos \theta = \frac{|-2|}{\sqrt{2} \sqrt{50}} = \frac{2}{\sqrt{2} \cdot 5\sqrt{2}} = \frac{2}{5 \cdot 2} = \frac{1}{5}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{5}\right)$.

(B) The direction ratios $(l, m, n)$ of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
Given points are $(1, 2, -1)$ and $(-1, 0, 1)$.
Calculating the differences:
$l = -1 - 1 = -2$
$m = 0 - 2 = -2$
$n = 1 - (-1) = 2$
Thus,the direction ratios are proportional to $(-2, -2, 2)$.
Dividing by $-2$,we get the ratios $(1, 1, -1)$.
Therefore,$(l, m, n) = (1, 1, -1)$.

(A) Let the first line be $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = k_1$. Then $x = 5k_1 + 4, y = 2k_1 + 1, z = k_1$.
Let the second line be $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = k_2$. Then $x = 2k_2 + 1, y = 3k_2 + 2, z = 4k_2 + 3$.
For the point of intersection,we equate the coordinates:
$5k_1 + 4 = 2k_2 + 1 \implies 5k_1 - 2k_2 = -3$ (Equation $1$)
$2k_1 + 1 = 3k_2 + 2 \implies 2k_1 - 3k_2 = 1$ (Equation $2$)
Solving these equations: Multiply Equation $1$ by $3$ and Equation $2$ by $2$:
$15k_1 - 6k_2 = -9$
$4k_1 - 6k_2 = 2$
Subtracting the equations: $11k_1 = -11 \implies k_1 = -1$.
Substituting $k_1 = -1$ into the first line coordinates: $x = 5(-1) + 4 = -1, y = 2(-1) + 1 = -1, z = -1$.
Thus,the point of intersection is $(-1, -1, -1)$.

(A) The given equations of the line are $x = ay + b$ and $z = cy + d$.
We can rewrite these equations as:
$x - ay = b$
$-cy + z = -d$
To find the direction ratios $(l, m, n)$,we express the variables in terms of a parameter $y$. Let $y = t$. Then:
$x = at + b$
$y = t$
$z = ct + d$
This is the parametric form of the line: $\frac{x - b}{a} = \frac{y - 0}{1} = \frac{z - d}{c} = t$.
The denominators represent the direction ratios of the line.
Therefore,the direction ratios are $(a, 1, c)$.

(B) line equally inclined to the axes has direction cosines $(l, m, n)$ such that $|l| = |m| = |n|$.
Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,which implies $l = \pm \frac{1}{\sqrt{3}}$.
Thus,the direction ratios $(a, b, c)$ can be taken as $(1, 1, 1)$ or $(1, -1, 1)$ etc.
The standard form of the line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting the point $(-3, 2, -4)$ and direction ratios $(1, 1, 1)$,we get $\frac{x - (-3)}{1} = \frac{y - 2}{1} = \frac{z - (-4)}{1}$.
This simplifies to $\frac{x + 3}{1} = \frac{y - 2}{1} = \frac{z + 4}{1}$,which is equivalent to $x + 3 = y - 2 = z + 4$.

(D) Let $A(0, 0, 0)$ be the origin and $B(-9, 4, 5)$ and $C(10, 0, -1)$ be the given points. Let $D$ be the foot of the perpendicular from $A$ to the line $BC$. Let $D$ divide $BC$ in the ratio $\lambda : 1$.
The coordinates of $D$ are given by the section formula:
$D = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{0\lambda + 4}{\lambda + 1}, \frac{-1\lambda + 5}{\lambda + 1} \right) = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{4}{\lambda + 1}, \frac{5 - \lambda}{\lambda + 1} \right)$.
The direction ratios of the line $BC$ are $(10 - (-9), 0 - 4, -1 - 5) = (19, -4, -6)$.
Since $AD \perp BC$,the dot product of vector $\vec{AD}$ and the direction vector of $BC$ must be zero.
$vec{AD} = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{4}{\lambda + 1}, \frac{5 - \lambda}{\lambda + 1} \right)$.
$19 \left( \frac{10\lambda - 9}{\lambda + 1} \right) - 4 \left( \frac{4}{\lambda + 1} \right) - 6 \left( \frac{5 - \lambda}{\lambda + 1} \right) = 0$.
$19(10\lambda - 9) - 16 - 6(5 - \lambda) = 0$.
$190\lambda - 171 - 16 - 30 + 6\lambda = 0$.
$196\lambda - 217 = 0 \Rightarrow \lambda = \frac{217}{196} = \frac{31}{28}$.
Substituting $\lambda = \frac{31}{28}$ into the coordinates of $D$:
$x = \frac{10(\frac{31}{28}) - 9}{\frac{31}{28} + 1} = \frac{310 - 252}{31 + 28} = \frac{58}{59}$.
$y = \frac{4}{\frac{31}{28} + 1} = \frac{4 \times 28}{59} = \frac{112}{59}$.
$z = \frac{5 - \frac{31}{28}}{\frac{31}{28} + 1} = \frac{140 - 31}{59} = \frac{109}{59}$.
The coordinates are $(\frac{58}{59}, \frac{112}{59}, \frac{109}{59})$,which is not among the options $A, B, C$. Therefore,the correct option is $D$.

(B) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.
Given points are $(a, b, c)$ and $(a - b, b - c, c - a)$.
Substituting these values into the formula:
$\frac{x - a}{(a - b) - a} = \frac{y - b}{(b - c) - b} = \frac{z - c}{(c - a) - c}$
Simplifying the denominators:
$\frac{x - a}{-b} = \frac{y - b}{-c} = \frac{z - c}{-a}$
Multiplying by $-1$ throughout:
$\frac{x - a}{b} = \frac{y - b}{c} = \frac{z - c}{a}$.

(D) The equation of a line passing through $(a, b, c)$ with direction ratios $(l, m, n)$ is given by $\frac{x - a}{l} = \frac{y - b}{m} = \frac{z - c}{n}$.
Since the line is parallel to the $z$-axis,its direction ratios are proportional to those of the $z$-axis,which are $(0, 0, 1)$.
Thus,we have $l = 0, m = 0$,and $n = 1$.
Substituting these values into the standard equation,we get $\frac{x - a}{0} = \frac{y - b}{0} = \frac{z - c}{1}$.

(B) Let the line be $L: \frac{x - 1}{2} = \frac{y}{9} = \frac{z}{5} = \lambda$.
Any point $Q$ on the line is given by $(2\lambda + 1, 9\lambda, 5\lambda)$.
The direction ratios of the line $PQ$ are $(2\lambda + 1 - 5, 9\lambda - 4, 5\lambda + 1) = (2\lambda - 4, 9\lambda - 4, 5\lambda + 1)$.
Since $PQ$ is perpendicular to the line $L$ with direction ratios $(2, 9, 5)$,their dot product is zero:
$2(2\lambda - 4) + 9(9\lambda - 4) + 5(5\lambda + 1) = 0$.
$4\lambda - 8 + 81\lambda - 36 + 25\lambda + 5 = 0$.
$110\lambda - 39 = 0 \implies \lambda = \frac{39}{110}$.
The coordinates of $Q$ are $(2(\frac{39}{110}) + 1, 9(\frac{39}{110}), 5(\frac{39}{110})) = (\frac{188}{110}, \frac{351}{110}, \frac{195}{110})$.
The length $PQ = \sqrt{(2\lambda - 4)^2 + (9\lambda - 4)^2 + (5\lambda + 1)^2}$.
Substituting $\lambda = \frac{39}{110}$,we get $PQ = \sqrt{\frac{2109}{110}}$.

(C) Let the given line be $L: \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = \lambda$.
Any point $P$ on the line is given by $P = (3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$.
Let the given point be $A = (1, 2, 3)$.
The vector $\vec{AP} = (3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4)$.
The direction vector of the line is $\vec{v} = (3, 2, -2)$.
Since $AP$ is perpendicular to the line,$\vec{AP} \cdot \vec{v} = 0$.
$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$.
$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$.
$17\lambda + 17 = 0 \implies \lambda = -1$.
Substituting $\lambda = -1$ into $P$,we get $P = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9)$.
The length of the perpendicular is the distance $AP = \sqrt{(3 - 1)^2 + (5 - 2)^2 + (9 - 3)^2}$.
$AP = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

(B) Given relations are $l + m + n = 0$ and $2lm + 2nl - mn = 0$.
From the first equation,$n = -(l + m)$.
Substituting this into the second equation: $2lm + 2(l + m)(-(l + m)) - m(-(l + m)) = 0$.
$2lm - 2(l^2 + 2lm + m^2) + m(l + m) = 0$.
$2lm - 2l^2 - 4lm - 2m^2 + ml + m^2 = 0$.
$-2l^2 - lm - m^2 = 0$,which simplifies to $2l^2 + lm + m^2 = 0$.
Wait,let us re-evaluate: $2lm + 2nl - mn = 0$.
Substituting $n = -(l + m)$: $2lm + 2l(-(l + m)) - m(-(l + m)) = 0$.
$2lm - 2l^2 - 2lm + ml + m^2 = 0$.
$-2l^2 + lm + m^2 = 0$,or $2l^2 - lm - m^2 = 0$.
Factoring gives $(2l + m)(l - m) = 0$.
Case $1$: $2l + m = 0 \Rightarrow m = -2l$. Then $n = -(l - 2l) = l$.
Direction ratios $(l, m, n)$ are $(l, -2l, l)$,i.e.,$(1, -2, 1)$.
Case $2$: $l - m = 0 \Rightarrow m = l$. Then $n = -(l + l) = -2l$.
Direction ratios $(l, m, n)$ are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Let $\theta$ be the angle between the lines.
$\cos \theta = \frac{|(1)(1) + (-2)(1) + (1)(-2)|}{\sqrt{1^2 + (-2)^2 + 1^2} \sqrt{1^2 + 1^2 + (-2)^2}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$.
Since the angle between lines is usually taken as acute,$\cos \theta = 1/2 \Rightarrow \theta = \pi/3$.
However,if we consider the angle between the vectors,$\cos \theta = \frac{-3}{6} = -1/2 \Rightarrow \theta = 2\pi/3$.

(C) Let the given point be $P(2, 4, -1)$ and the line be $L: \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda$.
Any point $Q$ on the line is given by $Q(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.
The vector $\vec{PQ} = (\lambda - 5 - 2, 4\lambda - 3 - 4, -9\lambda + 6 - (-1)) = (\lambda - 7, 4\lambda - 7, -9\lambda + 7)$.
The direction vector of the line is $\vec{v} = (1, 4, -9)$.
Since $PQ$ is perpendicular to the line,$\vec{PQ} \cdot \vec{v} = 0$.
$1(\lambda - 7) + 4(4\lambda - 7) - 9(-9\lambda + 7) = 0$.
$\lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0$.
$98\lambda - 98 = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ in $Q$,we get $Q(1 - 5, 4(1) - 3, -9(1) + 6) = Q(-4, 1, -3)$.
The distance $PQ = \sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2} = \sqrt{(-6)^2 + (-3)^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.

(D) The direction ratios of the first line are $\vec{a_1} = (2, 2, -1)$.
The direction ratios of the second line are $\vec{a_2} = (1, 2, 2)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values,we get $\cos \theta = \frac{|(2)(1) + (2)(2) + (-1)(2)|}{\sqrt{2^2 + 2^2 + (-1)^2} \sqrt{1^2 + 2^2 + 2^2}}$.
$\cos \theta = \frac{|2 + 4 - 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}} = \frac{4}{\sqrt{9} \sqrt{9}} = \frac{4}{3 \times 3} = \frac{4}{9}$.
Therefore,$\theta = \cos^{-1}\left(\frac{4}{9}\right)$.

(D) The direction ratios of the first line are $\vec{b_1} = (1, 2, 3)$.
The direction ratios of the second line are $\vec{b_2} = (2, 2, -2)$.
Both lines pass through the point $(1, 2, 3)$,so they intersect at this point.
To check the angle between them,we calculate the dot product of the direction vectors:
$\vec{b_1} \cdot \vec{b_2} = (1)(2) + (2)(2) + (3)(-2) = 2 + 4 - 6 = 0$.
Since the dot product is $0$,the lines are perpendicular to each other.
Therefore,the lines are intersecting at a right angle.

(B) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.
Given points are $(3, 2, 4)$ and $(4, 5, 2)$.
Substituting these values into the formula:
$\frac{x - 3}{4 - 3} = \frac{y - 2}{5 - 2} = \frac{z - 4}{2 - 4}$
$\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z - 4}{-2}$.
Thus,the correct option is $B$.

(C) The direction ratios of the first line are $\vec{b_1} = (1, 2, 3)$.
The direction ratios of the second line are $\vec{b_2} = (3, -2, 1)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the values:
$\cos \theta = \frac{|(1)(3) + (2)(-2) + (3)(1)|}{\sqrt{1^2 + 2^2 + 3^2} \sqrt{3^2 + (-2)^2 + 1^2}}$
$\cos \theta = \frac{|3 - 4 + 3|}{\sqrt{1 + 4 + 9} \sqrt{9 + 4 + 1}}$
$\cos \theta = \frac{2}{\sqrt{14} \sqrt{14}} = \frac{2}{14} = \frac{1}{7}$
Therefore,$\theta = \cos^{-1}\left(\frac{1}{7}\right)$.

(A) The direction ratios of the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ are given by $(a_1, b_1, c_1) = (-3 - 2, 1 - 1, 7 - (-3)) = (-5, 0, 10)$.
The direction ratios of the line parallel to $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$ are $(a_2, b_2, c_2) = (3, 4, 5)$.
The cosine of the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values,we get $\cos \theta = \frac{|(-5)(3) + (0)(4) + (10)(5)|}{\sqrt{(-5)^2 + 0^2 + 10^2} \sqrt{3^2 + 4^2 + 5^2}}$.
$\cos \theta = \frac{|-15 + 0 + 50|}{\sqrt{25 + 0 + 100} \sqrt{9 + 16 + 25}} = \frac{35}{\sqrt{125} \sqrt{50}}$.
$\cos \theta = \frac{35}{5\sqrt{5} \times 5\sqrt{2}} = \frac{35}{25\sqrt{10}} = \frac{7}{5\sqrt{10}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{7}{5\sqrt{10}}\right)$.

(D) The direction ratios of the first line are $a_1 = 2, b_1 = 5, c_1 = 4$.
The direction ratios of the second line are $a_2 = 1, b_2 = 2, c_2 = -3$.
Let $\theta$ be the angle between the lines. The formula for the angle is $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\cos \theta = \frac{|(2)(1) + (5)(2) + (4)(-3)|}{\sqrt{2^2 + 5^2 + 4^2} \sqrt{1^2 + 2^2 + (-3)^2}}$.
$\cos \theta = \frac{|2 + 10 - 12|}{\sqrt{4 + 25 + 16} \sqrt{1 + 4 + 9}} = \frac{|0|}{\sqrt{45} \sqrt{14}} = 0$.
Since $\cos \theta = 0$,we have $\theta = 90^o$.

(A) Given equations are:
$3lm - 4ln + mn = 0$ --- $(1)$
$l + 2m + 3n = 0$ --- $(2)$
From $(2)$,$l = -2m - 3n$. Substituting this into $(1)$:
$3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0$
$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0$
$m^2 = 2n^2 \implies m = \pm \sqrt{2}n$
Case $1$: If $m = \sqrt{2}n$,then $l = -2(\sqrt{2}n) - 3n = -(2\sqrt{2} + 3)n$. Direction ratios are $(-(2\sqrt{2} + 3), \sqrt{2}, 1)$.
Case $2$: If $m = -\sqrt{2}n$,then $l = -2(-\sqrt{2}n) - 3n = (2\sqrt{2} - 3)n$. Direction ratios are $((2\sqrt{2} - 3), -\sqrt{2}, 1)$.
Let the direction ratios be $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$.
$a_1 a_2 + b_1 b_2 + c_1 c_2 = (-(2\sqrt{2} + 3))(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$
$= -((2\sqrt{2})^2 - 3^2) - 2 + 1 = -(8 - 9) - 1 = 1 - 1 = 0$.
Since the dot product of the direction vectors is $0$,the lines are perpendicular.
Thus,the angle between the lines is $\pi /2$.

(C) The $x$-axis passes through the origin $(0, 0, 0)$.
Its direction ratios are $(1, 0, 0)$ because it is parallel to the $x$-axis.
The symmetric form of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a, b, c) = (1, 0, 0)$,we get $\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}$.
This simplifies to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$.

(D) The equation of the given line is $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$.
The direction ratios of this line are given by the denominators,which are $\vec{v} = (3, 1, 0)$.
The direction ratios of the $z$-axis are $\vec{k} = (0, 0, 1)$.
To check if the line is perpendicular to the $z$-axis,we calculate the dot product of the direction vectors:
$\vec{v} \cdot \vec{k} = (3)(0) + (1)(0) + (0)(1) = 0 + 0 + 0 = 0$.
Since the dot product is $0$,the line is perpendicular to the $z$-axis.

(A) Let the direction ratios of the first line be $\vec{a} = (2, 2, 1)$.
The second line joins the points $P(3, 1, 4)$ and $Q(7, 2, 12)$.
The direction ratios of the second line are $\vec{b} = (7-3, 2-1, 12-4) = (4, 1, 8)$.
The cosine of the angle $\theta$ between the two lines is given by the formula:
$\cos \theta = \frac{|a_1b_1 + a_2b_2 + a_3b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$
$\cos \theta = \frac{|2 \times 4 + 2 \times 1 + 1 \times 8|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}$
$\cos \theta = \frac{|8 + 2 + 8|}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}}$
$\cos \theta = \frac{18}{\sqrt{9} \sqrt{81}} = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$
Therefore,$\theta = \cos^{-1}(2/3)$.

(A) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.
Given points are $(x_1, y_1, z_1) = (4, -5, -2)$ and $(x_2, y_2, z_2) = (-1, 5, 3)$.
Substituting these values into the formula:
$\frac{x - 4}{-1 - 4} = \frac{y - (-5)}{5 - (-5)} = \frac{z - (-2)}{3 - (-2)}$
Simplifying the denominators:
$\frac{x - 4}{-5} = \frac{y + 5}{10} = \frac{z + 2}{5}$
Dividing the denominators by $-5$:
$\frac{x - 4}{1} = \frac{y + 5}{-2} = \frac{z + 2}{-1}$.
Thus,the correct option is $A$.

(A) Let the given lines be:
$\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1} = r_1$
$\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4} = r_2$
From the first line,any point is $(3r_1 + 5, -r_1 + 7, r_1 - 2)$.
From the second line,any point is $(-36r_2 - 3, 2r_2 + 3, 4r_2 + 6)$.
Equating the coordinates:
$3r_1 + 5 = -36r_2 - 3 \implies 3r_1 + 36r_2 = -8$ (Eq. $1$)
$-r_1 + 7 = 2r_2 + 3 \implies r_1 + 2r_2 = 4 \implies r_1 = 4 - 2r_2$ (Eq. $2$)
Substitute $r_1$ in Eq. $1$:
$3(4 - 2r_2) + 36r_2 = -8$
$12 - 6r_2 + 36r_2 = -8$
$30r_2 = -20 \implies r_2 = -2/3$
Now,$r_1 = 4 - 2(-2/3) = 4 + 4/3 = 16/3$.
Substituting $r_1$ into the first line coordinates:
$x = 3(16/3) + 5 = 16 + 5 = 21$
$y = -(16/3) + 7 = (-16 + 21)/3 = 5/3$
$z = (16/3) - 2 = (16 - 6)/3 = 10/3$
The point of intersection is $(21, 5/3, 10/3)$.

(D) The given lines are $2x = 3y = -z$ and $6x = -y = -4z$.
First,we write the lines in symmetric form $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$.
For the first line: $2x = 3y = -z \Rightarrow \frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{-1}$. Multiplying by $6$,we get $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$. The direction ratios are $\vec{v_1} = (3, 2, -6)$.
For the second line: $6x = -y = -4z \Rightarrow \frac{x}{1/6} = \frac{y}{-1} = \frac{z}{-1/4}$. Multiplying by $12$,we get $\frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}$. The direction ratios are $\vec{v_2} = (2, -12, -3)$.
Two lines are perpendicular if the dot product of their direction vectors is zero: $\vec{v_1} \cdot \vec{v_2} = (3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.
Since the dot product is $0$,the angle between the lines is $90^o$.

(C) Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
For the given lines,the direction ratios are $(-3, 2k, 2)$ and $(3k, 1, -5)$.
Applying the condition for perpendicularity:
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k - 10 = 0$
$-7k = 10$
$k = -\frac{10}{7}$.

(D) The given line is $1 - x = \frac{y}{2} = \frac{1}{3}(1 + z)$.
Rewriting the line in standard form: $\frac{x - 1}{-1} = \frac{y - 0}{2} = \frac{z - (-1)}{3}$.
Let the point $P = (2, 3, 4)$ and the line pass through $A = (1, 0, -1)$ with direction vector $\vec{v} = (-1, 2, 3)$.
The vector $\vec{AP} = (2-1, 3-0, 4-(-1)) = (1, 3, 5)$.
The distance $d$ of point $P$ from the line is given by $d = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$.
$\vec{AP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 5 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}(9 - 10) - \hat{j}(3 + 5) + \hat{k}(2 + 3) = -\hat{i} - 8\hat{j} + 5\hat{k}$.
$|\vec{AP} \times \vec{v}| = \sqrt{(-1)^2 + (-8)^2 + 5^2} = \sqrt{1 + 64 + 25} = \sqrt{90} = 3\sqrt{10}$.
$|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
$d = \frac{3\sqrt{10}}{\sqrt{14}} = \frac{3\sqrt{5}}{\sqrt{7}} = \frac{3\sqrt{35}}{7} = \frac{3}{7}\sqrt{35}$.

(B) The direction ratios of the first line are $\vec{a_1} = (2, 5, -3)$.
The direction ratios of the second line are $\vec{a_2} = (-1, 8, 4)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
$\cos \theta = \frac{|2(-1) + 5(8) + (-3)(4)|}{\sqrt{2^2 + 5^2 + (-3)^2} \sqrt{(-1)^2 + 8^2 + 4^2}}$
$\cos \theta = \frac{|-2 + 40 - 12|}{\sqrt{4 + 25 + 9} \sqrt{1 + 64 + 16}}$
$\cos \theta = \frac{|26|}{\sqrt{38} \sqrt{81}} = \frac{26}{9\sqrt{38}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)$.

(A) The given equation is $y^2 + z^2 = 0$.
Since $y^2 \ge 0$ and $z^2 \ge 0$ for all real numbers $y$ and $z$,the sum $y^2 + z^2$ can be zero if and only if both $y = 0$ and $z = 0$ simultaneously.
In three-dimensional space,the set of points $(x, y, z)$ satisfying $y = 0$ and $z = 0$ corresponds to all points where the $y$-coordinate and $z$-coordinate are zero,regardless of the value of $x$.
This set of points forms the $x$-axis.
Therefore,the correct option is $A$.

(C) The equation of the line passing through $P(3, 4, 1)$ and $Q(5, 1, 6)$ is given by $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1} = k$.
This simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.
Any point on this line is $(2k+3, -3k+4, 5k+1)$.
Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$.
Therefore,$5k+1 = 0$,which gives $k = -\frac{1}{5}$.
Substituting $k = -\frac{1}{5}$ into the coordinates:
$x = 2(-\frac{1}{5}) + 3 = -\frac{2}{5} + \frac{15}{5} = \frac{13}{5}$.
$y = -3(-\frac{1}{5}) + 4 = \frac{3}{5} + \frac{20}{5} = \frac{23}{5}$.
$z = 0$.
Thus,the point is $(\frac{13}{5}, \frac{23}{5}, 0)$.

(A) The equation of the line passing through points $(x_1, y_1, z_1) = (3, 5, -7)$ and $(x_2, y_2, z_2) = (-2, 1, 8)$ is given by:
$\frac{x - 3}{-2 - 3} = \frac{y - 5}{1 - 5} = \frac{z - (-7)}{8 - (-7)}$
$\frac{x - 3}{-5} = \frac{y - 5}{-4} = \frac{z + 7}{15} = K$
Any point on this line is given by $(x, y, z) = (-5K + 3, -4K + 5, 15K - 7)$.
Since the line meets the $yz$-plane,the $x$-coordinate must be $0$.
$-5K + 3 = 0 \Rightarrow K = \frac{3}{5}$.
Substituting $K = \frac{3}{5}$ into the coordinates:
$x = -5(\frac{3}{5}) + 3 = 0$
$y = -4(\frac{3}{5}) + 5 = -\frac{12}{5} + \frac{25}{5} = \frac{13}{5}$
$z = 15(\frac{3}{5}) - 7 = 9 - 7 = 2$
Thus,the point of intersection is $\left( 0, \frac{13}{5}, 2 \right)$.

(A) Let the direction ratios of the required line be $(l, m, n)$.
Since the line is perpendicular to the lines with direction ratios $(3, -16, 7)$ and $(3, 8, -5)$,we have:
$3l - 16m + 7n = 0$ $(i)$
$3l + 8m - 5n = 0$ (ii)
Subtracting (ii) from $(i)$,we get: $-24m + 12n = 0 \implies 24m = 12n \implies n = 2m$.
Substituting $n = 2m$ in $(i)$: $3l - 16m + 7(2m) = 0 \implies 3l - 16m + 14m = 0 \implies 3l = 2m \implies l = \frac{2}{3}m$.
Taking $m = 3$,we get $l = 2$ and $n = 6$.
Thus,the direction ratios are $(2, 3, 6)$.
The equation of the line passing through $(1, 2, -4)$ with direction ratios $(2, 3, 6)$ is $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$.

(A) Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.
From the first equation,$n = -\frac{al + bm}{c}$.
Substituting this into the second equation: $fm(-\frac{al + bm}{c}) + gl(-\frac{al + bm}{c}) + hlm = 0$.
Multiplying by $-c$,we get $fm(al + bm) + gl(al + bm) - chlm = 0$.
$aflm + bfm^2 + agl^2 + bglm - chlm = 0$.
$agl^2 + (af + bg - ch)lm + bfm^2 = 0$.
Dividing by $m^2$,we get $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0$.
Let the roots be $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. Then $\frac{l_1 l_2}{m_1 m_2} = \frac{bf}{ag}$,which implies $\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b}$.
Similarly,by eliminating $l$,we get $\frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c}$.
Thus,$\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c} = k$.
For perpendicular lines,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Substituting the values,$k(\frac{f}{a} + \frac{g}{b} + \frac{h}{c}) = 0$.
Since $k \neq 0$,we have $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

(B) Let any point on the first line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = \lambda$ be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$.
Let any point on the second line $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = \mu$ be $(\mu + 3, 2\mu + k, \mu)$.
The lines intersect if there exist $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda + 1 = \mu + 3 \implies 2\lambda - \mu = 2$ $(i)$
$3\lambda - 1 = 2\mu + k \implies 3\lambda - 2\mu = k + 1$ (ii)
$4\lambda + 1 = \mu \implies 4\lambda - \mu = -1$ (iii)
Subtracting $(i)$ from (iii):
$(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$
$2\lambda = -3 \implies \lambda = -\frac{3}{2}$.
Substitute $\lambda = -\frac{3}{2}$ into (iii):
$4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$.
Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into (ii):
$3(-\frac{3}{2}) - 2(-5) = k + 1$
$-\frac{9}{2} + 10 = k + 1$
$k = -\frac{9}{2} + 9 = \frac{9}{2}$.

(A) Eliminating $t$ from the given equations,we get the equation of the path:
$\frac{x}{2} = \frac{y}{-4} = \frac{z}{4}$ or $\frac{x}{1} = \frac{y}{-2} = \frac{z}{2}$.
This represents a straight line passing through the origin.
For $t = 10 \text{ s}$,the coordinates are:
$x = 2(10) = 20 \text{ km}$,
$y = -4(10) = -40 \text{ km}$,
$z = 4(10) = 40 \text{ km}$.
The distance from the origin $O(0, 0, 0)$ is given by:
$d = \sqrt{x^2 + y^2 + z^2} = \sqrt{20^2 + (-40)^2 + 40^2}$
$d = \sqrt{400 + 1600 + 1600} = \sqrt{3600} = 60 \text{ km}$.

(B) Let $A = (1, 0, 3)$,$B = (4, 7, 1)$,and $C = (3, 5, 3)$. Let $P$ be the foot of the perpendicular from $A$ to the line $BC$.
The equation of the line passing through $B$ and $C$ is given by $\vec{r} = \vec{b} + \lambda(\vec{c} - \vec{b})$.
Direction ratios of $BC$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.
So,the line $BC$ is $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = k$.
Any point $P$ on the line $BC$ is given by $P = (4-k, 7-2k, 1+2k)$.
The vector $\vec{AP} = (4-k-1, 7-2k-0, 1+2k-3) = (3-k, 7-2k, 2k-2)$.
Since $AP \perp BC$,the dot product of $\vec{AP}$ and the direction vector of $BC$ (which is $(-1, -2, 2)$) must be zero.
$(3-k)(-1) + (7-2k)(-2) + (2k-2)(2) = 0$
$-3 + k - 14 + 4k + 4k - 4 = 0$
$9k - 21 = 0 \Rightarrow 9k = 21 \Rightarrow k = \frac{21}{9} = \frac{7}{3}$.
Substituting $k = \frac{7}{3}$ into the coordinates of $P$:
$x = 4 - \frac{7}{3} = \frac{12-7}{3} = \frac{5}{3}$
$y = 7 - 2(\frac{7}{3}) = \frac{21-14}{3} = \frac{7}{3}$
$z = 1 + 2(\frac{7}{3}) = \frac{3+14}{3} = \frac{17}{3}$
Thus,the foot of the perpendicular is $\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$.

(A) Given equations are $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$.
From the second equation,$l = -2m - 3n$.
Substituting this into the first equation: $3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0$.
$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$.
$-6m^2 + 12n^2 = 0$,which implies $m^2 = 2n^2$,so $m = \pm \sqrt{2}n$.
Case $1$: $m = \sqrt{2}n$. Then $l = -2(\sqrt{2}n) - 3n = -(2\sqrt{2} + 3)n$. Direction ratios are $(-(2\sqrt{2} + 3), \sqrt{2}, 1)$.
Case $2$: $m = -\sqrt{2}n$. Then $l = -2(-\sqrt{2}n) - 3n = (2\sqrt{2} - 3)n$. Direction ratios are $((2\sqrt{2} - 3), -\sqrt{2}, 1)$.
Let the direction ratios be $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$.
$a_1 a_2 + b_1 b_2 + c_1 c_2 = (-(2\sqrt{2} + 3))(2\sqrt{2} - 3) + (\sqrt{2})(-\sqrt{2}) + (1)(1)$.
$= -((2\sqrt{2})^2 - 3^2) - 2 + 1 = -(8 - 9) - 1 = -(-1) - 1 = 1 - 1 = 0$.
Since the dot product of the direction ratios is $0$,the lines are perpendicular.
Therefore,the angle between them is $\pi /2$.